Prove that ##\lambda## or ##-\lambda## is an eigenvalue for ##T##.

[Hall]

Homework Statement

If $T: V \to V$ has the property that $T^2$ has a non-negative eigenvalue $\lambda^2$ , prove that at least one
of $\lambda$ or $-\lambda$ is an eigenvalue for T. [Hint: $T^2 - \lambda^2 = (T + \lambda I)(T - \lambda I)$ .]

Relevant Equations

$T(x)= \lambda x$

The statement " If $T: V \to V$ has the property that $T^2$ has a non-negative eigenvalue $\lambda^2$", means that there exists an $x$ in $V$ such that $T^2 (x) = \lambda^2 x$.

If $T(x) = \mu x$, we've have
$$
T [T(x)]= T ( \mu x)$$
$$
T^2 (x) = \mu^2 x$$
$$
\lambda ^2 = \mu ^2 \implies \mu = \lambda ~or~ -\lambda$$.

Is my solution correct? The only thing which is quite not very well is that I assumed that there exists an eigenvalue $\mu$ and an eigenvector $x$ in $T$.

What the hint in the question wants me to do?

 

[fresh_42]

Homework Statement:: If $T: V \to V$ has the property that $T^2$ has a non-negative eigenvalue $\lambda^2$ , prove that at least one
of $\lambda$ or $-\lambda$ is an eigenvalue for T. [Hint: $T^2 - \lambda^2 = (T + \lambda I)(T - \lambda I)$ .]
Relevant Equations:: $T(x)= \lambda x$

The statement " If $T: V \to V$ has the property that $T^2$ has a non-negative eigenvalue $\lambda^2$", means that there exists an $x$ in $V$ such that $T^2 (x) = \lambda^2 x$.

If $T(x) = \mu x$, we've have
$$
T [T(x)]= T ( \mu x)$$
$$
T^2 (x) = \mu^2 x$$
$$
\lambda ^2 = \mu ^2 \implies \mu = \lambda ~or~ -\lambda$$.

Is my solution correct?

No.

The only thing which is quite not very well is that I assumed that there exists an eigenvalue $\mu$ and an eigenvector $x$ in $T$.

You assumed that the eigenvector $x$ to the eigenvalue $\mu$ is the same vector as the eigenvector to the eigenvalue $\lambda^2$ of $T^2.$ I do not see that this is necessarily true. And what if $T$ hasn't any real eigenvalues at all?

What the hint in the question wants me to do?

You started correctly: There is a vector $x\neq 0$ such that $T^2(x)=\lambda^2 x.$

Now use the hint: $(T^2-\lambda^2I)(x)=(T+\lambda )(T-\lambda )(x)= (T-\lambda )(T+\lambda )(x)=0$
and consider the case $T(x)\neq \pm \lambda x$ since otherwise we are done.

Can you find an eigenvector to $\lambda$ or $-\lambda$ anyway?

 

[fresh_42]

A good habit to avoid mistakes like this is by the notation: Always note dependences by indices or similar.

Example 1 (epsilontic):

For all $\varepsilon >0$ exists an $N$ such that $|a_n-L|<\varepsilon$ for all $n>N$.

should actually be

For all $\varepsilon >0$ exists an $N(\varepsilon )$ such that $|a_n-L|<\varepsilon$ for all $n>N(\varepsilon )$.

Example 2 (eigenvectors):

Assume $x$ is an eigenvector of $T^2$ to the eigenvalue $\lambda^2,$ i.e. $T^2(x)=\lambda^2x.$

should actually be

Assume $x_{\nu}$ is an eigenvector of $T^2$ to the eigenvalue $\nu:=\lambda^2,$ i.e. $T^2(x_{\nu})=\nu x_{\nu}=\lambda^2x_{\nu}.$

 

[Office_Shredder]

I think a nice example (not quite right since the question is assuming real numbers) here is a rotation matrix in 2 dimension $T(x,y)=(-y,x)$. $T^2$ is negative the identity so every vector is an eigenvector. But the eigenvector of $T$ is not obvious. It turns out $-i$ is an eigenvalue in $\mathbb{C}^2$

 

[Delta2]

Hmmm, I can't solve this unless it is given that $ker(T)=\{0_V\}$.

 

[Hall]

No.

You assumed that the eigenvector $x$ to the eigenvalue $\mu$ is the same vector as the eigenvector to the eigenvalue $\lambda^2$ of $T^2.$ I do not see that this is necessarily true. And what if $T$ hasn't any real eigenvalues at all?You started correctly: There is a vector $x\neq 0$ such that $T^2(x)=\lambda^2 x.$

Now use the hint: $(T^2-\lambda^2I)(x)=(T+\lambda )(T-\lambda )(x)= (T-\lambda )(T+\lambda )(x)=0$
and consider the case $T(x)\neq \pm \lambda x$ since otherwise we are done.

Can you find an eigenvector to $\lambda$ or $-\lambda$ anyway?

$[T^2 - (\lambda I)^2] (x) = (T+ \lambda I) (T- \lambda I) (x)$
$T^2 (x) - \lambda^2 I(x) = (T + \lambda I) [T(x) - \lambda I(x)]$
$\lambda^2 x - \lambda ^2 x = (T +\lambda I) (T(x) - \lambda x)$
$(T +\lambda I) (T(x) - \lambda x) = 0$
But without being given that $(T + \lambda I)$ is one-to-one, we cannot equate thus
$T(x) - \lambda x = 0$
$T(x) = \lambda x$

(Changing the order in the very first equation to $(T- \lambda I) (T+ \lambda I)$ would give us $T(x) = -\lambda x$).

 

[fresh_42]

$[T^2 - (\lambda I)^2] (x) = (T+ \lambda I) (T- \lambda I) (x)$
$T^2 (x) - \lambda^2 I(x) = (T + \lambda I) [T(x) - \lambda I(x)]$
$\lambda^2 x - \lambda ^2 x = (T +\lambda I) (T(x) - \lambda x)$
$(T +\lambda I) (T(x) - \lambda x) = 0$
But without being given that $(T + \lambda I)$ is one-to-one, we cannot equate thus
$T(x) - \lambda x = 0$
$T(x) = \lambda x$

(Changing the order in the very first equation to $(T- \lambda I) (T+ \lambda I)$ would give us $T(x) = -\lambda x$).

Step by step. You are right, you cannot equate this.

We have $(T-\lambda I)(T+\lambda I)(x)=0$ and $T(x)\neq -\lambda x.$ (The other case is according.)
We need a vector $v$ such that $Tv=\lambda v$ and $v\neq 0.$

Can you see what $v$ is? It is already written there!

 

[Delta2]

I think me and @Hall did same mistake, we expected the $v$ of post #7 to be equal to the eigenvector $x$ of $T^2$, but $v$ can be different (can be the same though, it depends if $T$ is 1-1 it is the same ).

 

[Hall]

Step by step. You are right, you cannot equate this.

We have $(T-\lambda I)(T+\lambda I)(x)=0$ and $T(x)\neq -\lambda x.$ (The other case is according.)
We need a vector $v$ such that $Tv=\lambda v$ and $v\neq 0.$

Can you see what $v$ is? It is already written there!

$(T-\lambda I)(T+\lambda I)(x)=0$
$$T [ (T+ \lambda I) (x) ] - \lambda I [(T+ \lambda I)(x) ]= 0$$
$T [ (T+ \lambda I) (x) ] - \lambda [(T+ \lambda I)(x) ]= 0$
$$
T [ (T+ \lambda I) (x) ] = \lambda [(T+ \lambda I)(x) ]$$

Eigenvector = $(T+ \lambda I) (x)$, Eigenvalue= $\lambda$.
(We must state that $T(x) \neq -\lambda x$, else the eigenvector will be zero, which we don't want).
The second case:
$$
T^2 - (\lambda I)^2 = (T + \lambda I)(T- \lambda I) $$
$[T^2 - (\lambda I)^2](x) =[ (T +\lambda I)(T- \lambda I)] (x)$
$0 = T [ (T- \lambda I) (x)] + \lambda I [(T- \lambda I) (x)]$
$T [ (T- \lambda I) (x)] = - \lambda [(T- \lambda I) (x)]$

Eigenvector = $(T- \lambda I)(x)$, Eigenvalue = $-\lambda$.
(with the condition that $T(x) \neq \lambda x$)

 

[fresh_42]

$(T-\lambda I)(T+\lambda I)(x)=0$
$$T [ (T+ \lambda I) (x) ] - \lambda I [(T+ \lambda I)(x) ]= 0$$
$T [ (T+ \lambda I) (x) ] - \lambda [(T+ \lambda I)(x) ]= 0$
$$
T [ (T+ \lambda I) (x) ] = \lambda [(T+ \lambda I)(x) ]$$

Eigenvector = $(T+ \lambda I) (x)$, Eigenvalue= $\lambda$.
(We must state that $T(x) \neq -\lambda x$, else the eigenvector will be zero, which we don't want).
The second case:
$$
T^2 - (\lambda I)^2 = (T + \lambda I)(T- \lambda I) $$
$[T^2 - (\lambda I)^2](x) =[ (T +\lambda I)(T- \lambda I)] (x)$
$0 = T [ (T- \lambda I) (x)] + \lambda I [(T- \lambda I) (x)]$
$T [ (T- \lambda I) (x)] = - \lambda [(T- \lambda I) (x)]$

Eigenvector = $(T- \lambda I)(x)$, Eigenvalue = $-\lambda$.
(with the condition that $T(x) \neq \lambda x$)

Looks good. Now you only have to gather all possible cases: $Tx=\lambda x\, , \,Tx=-\lambda x\, , \,Tx\neq \lambda x\, , \,Tx\neq -\lambda x$ or reason with symmetry.

 

[Hall]

Looks good. Now you only have to gather all possible cases: $Tx=\lambda x\, , \,Tx=-\lambda x\, , \,Tx\neq \lambda x\, , \,Tx\neq -\lambda x$ or reason with symmetry.

Okay, we can do that. But why do we need to look at only those cases? If we're not done in Post #9, then we can consider many cases for $T(x)$.

 

[fresh_42]

Okay, we can do that. But why do we need to look at only those cases? If we're not done in Post #9, then we can consider many cases for $T(x)$.

These are all possible cases. If $Tx=\pm\lambda x$ then we are done. And $Tx\neq\pm\lambda x$ is what you did in post #9. You mentioned yourself that we need the vector to be different from zero. So the case if it is zero is left. It is only an especially easy case.